anonymous
  • anonymous
Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
First Differentiate the function to find the slope.
anonymous
  • anonymous
thank you but i dont know process. i am not sure about calculation, particulary differentiation
anonymous
  • anonymous
I mean i know the process

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zepdrix
  • zepdrix
You know the process, but you need assistance with the differentiation portion?
anonymous
  • anonymous
Yes, when it comes to both chain rule and products. I get really confused
zepdrix
  • zepdrix
\[\large 2(x^2+y^2)^2=25(x^2−y^2)\]So we'll be differentiating this `implicitly`, meaning, `as is`. No solving for y ahead of time. When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable. \(\large y'(-3,1)=m\). This is the m in our equation \(\large y=mx+b\). Hmm so let's see...
zepdrix
  • zepdrix
It appears we have no product rule which is nice.
anonymous
  • anonymous
OK, Y7=(-3,1)=m. right?
zepdrix
  • zepdrix
We should be able to get a numerical value for m, no x's or y's in our m. I'm not really sure what that Y7 is.
anonymous
  • anonymous
Ops my mistake
zepdrix
  • zepdrix
Taking the derivative of the left side gives us, `taking the derivative of the outermost function first, the square on the brackets`:\[\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}\]The blue is due to the chain rule. Understand that part?
anonymous
  • anonymous
yes, so far good.
zepdrix
  • zepdrix
\[\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}\]So we get something like this on the left side.
anonymous
  • anonymous
i lied, I know product rule, power rule etc, but i dont know what the chain rule is
zepdrix
  • zepdrix
lol :)
zepdrix
  • zepdrix
mmmm i always have trouble explaining that one. It's very unintuitive.
zepdrix
  • zepdrix
hmm
anonymous
  • anonymous
could you give me example of chain rule problem maybe?
zepdrix
  • zepdrix
The definition says this,\[\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)\]When you take the derivative of this, a function within a function, you ~Differentiate the outer function, f in this case. ~Then you make a copy of the inner function, and multiply by it's derivative. Here is a silly example. \[\large \frac{d}{dx}\left(x^2\right)^{4}\] In this example we're thinking of \(\large g(x)=x^2\). And \(\large f(x)=x^4\). If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), \(\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4\) will become, \(\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4\) which is \(\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4\). Applying the chain rule gives us,\[\large f'(g(x))=4\left(g(x)\right)^3 g'(x)\]See how the rule tells us to make a copy of the inner function and take it's derivative?\[\large f'(g(x))=4\left(x^2\right)^3 (x^2)'\]\[\large f'(g(x))=4\left(x^2\right)^3 (2x)\]
anonymous
  • anonymous
Oo.. I see, Thank you. It was very clear! really cool :)
zepdrix
  • zepdrix
If you were to simplify this down, you would get,\[\large f'(g(x))=8x^7\] Which we know is correct, because we could have easily approached this example without using the chain rule.\[\large y=(x^2)^4=x^8\]And we know the derivative will match what we got from the chain rule example.
zepdrix
  • zepdrix
It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol
anonymous
  • anonymous
no, you explain well. i always get confused and could not tell wether i should use chain or product.
zepdrix
  • zepdrix
So we applied the derivative of the left side,\[\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'\] right? :O
anonymous
  • anonymous
haha thats even more clear haha def, taking note
zepdrix
  • zepdrix
lol :d
anonymous
  • anonymous
could we back to the first question?
anonymous
  • anonymous
i took note
zepdrix
  • zepdrix
This one? :o \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]
anonymous
  • anonymous
yes, haha
anonymous
  • anonymous
I see the chain rule now, (x^2+y^2)^2?
zepdrix
  • zepdrix
\[\large 2(x^2+y^2)\]That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.\[\large 2(x^2+y^2)(x^2+y^2)'\]
zepdrix
  • zepdrix
\[\large 2(x^2+y^2)(2x+2y)\]So we should get somethingggggggg like this, yes?
anonymous
  • anonymous
:O why two in the fron of the bracket? I understand the (2x+2y) tho
zepdrix
  • zepdrix
You applied the power rule to the outer function \(\large \left(\qquad \right)^2\)
anonymous
  • anonymous
power rule and chain double? :O
zepdrix
  • zepdrix
Power rule the outside, make a copy of the inside, take derivative of copy.
anonymous
  • anonymous
Ok, i understand
zepdrix
  • zepdrix
Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.
zepdrix
  • zepdrix
\[\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})\]See this blue term? We took the derivative of a variable that is not x, namely y, so a y' popped out.
anonymous
  • anonymous
mmmmmm.... Ok. I think :/
anonymous
  • anonymous
so whenever its chain rule i put y' in the end?
zepdrix
  • zepdrix
No. Let's not think of that portion as chain rule, I think that's just going to confuse you. Whenever we take the derivative of y, we will multiply by y'. We don't throw it on the end, we always attach it to the y that we're taking the derivative of.
anonymous
  • anonymous
right, Ok. i will remember that by my heart.
zepdrix
  • zepdrix
lol c:
anonymous
  • anonymous
haha
anonymous
  • anonymous
whats nxt stp?
zepdrix
  • zepdrix
Try taking the derivative of the right side, tell me what you get.
anonymous
  • anonymous
Ok!
zepdrix
  • zepdrix
Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.
anonymous
  • anonymous
25(2x-2y)?
zepdrix
  • zepdrix
close, but you took the derivative of a y, you need to multiply that part by y'
anonymous
  • anonymous
25(2x-2yy')?
anonymous
  • anonymous
its still funny how i have to put y' in the end. is it for implicit derviation?
zepdrix
  • zepdrix
yes. If we had an equation like this,\[\large y=x^2\]The derivative gives us,\[\large y'=2x\]Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?\[\large y^2=x^2\]Taking the derivative,\[\large 2y=2x\] Hmm we don't have a derivative term do we?
anonymous
  • anonymous
Ok, thats cool
anonymous
  • anonymous
so 2(x^2+y~2)(2x+2y)=25(2x-2yy')
zepdrix
  • zepdrix
Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?
zepdrix
  • zepdrix
2(x^2+y^2)^2=25(x^2−y^2)
zepdrix
  • zepdrix
See that 2 at the front? That would give us an extra factor of 2 in our derivative.
anonymous
  • anonymous
the question equation is 2(x^2+y^2)^2=25(x^2-y^2)
anonymous
  • anonymous
so is it 2*2?
zepdrix
  • zepdrix
yes
anonymous
  • anonymous
\[4(x^2+y^2)(2x+2y)=25(2x-2yy')\]
zepdrix
  • zepdrix
\[\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x-2yy')\]Woops you missed a y'.
anonymous
  • anonymous
ops! haha thats derivative of the first equation?
zepdrix
  • zepdrix
yes. We took the derivative of y on each side, so each y will get a y'
anonymous
  • anonymous
Ok,
zepdrix
  • zepdrix
Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate \(\large (-3,1\;)\) and THEN solve for y'.
anonymous
  • anonymous
plug both number into the equation?
zepdrix
  • zepdrix
yes, -3 for all the x's. 1 in for all the y's.
zepdrix
  • zepdrix
and don't plug anything in for y'. That's what we'll be solving for.

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